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Question Number 145256 by Gbenga last updated on 03/Jul/21

$$x^{x^x}={\left(\frac{1}{2}\right)}^{\sqrt{2}}$$$$\mathbf{find}\mathbf{x}$$

Commented by justtry last updated on 04/Jul/21

$$ithinkthereisnotx\in \mathbb{R}tofindit.$$$$$$

Answered by mr W last updated on 04/Jul/21

$${\left(\frac{1}{2}\right)}^{\sqrt{2}}$$$$={\left(\frac{1}{4}\right)}^{\frac{\sqrt{2}}{2}}$$$$={\left(\frac{1}{4}\right)}^{2^{-\frac{1}{2}}}$$$$={\left(\frac{1}{4}\right)}^{4^{-\frac{1}{4}}}$$$$={\left(\frac{1}{4}\right)}^{{\left(\frac{1}{4}\right)}^{\frac{1}{4}}}$$$$x^{x^x}={\left(\frac{1}{4}\right)}^{{\left(\frac{1}{4}\right)}^{\frac{1}{4}}}$$$$\Rightarrow x=\frac{1}{4}$$

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